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Cathode rays. How are cathode rays produced in a discharge tube? What is the principle of thermionic emission? What is the work done on an electron accelerated through a p.d .?. Who invented the CRT?. - PowerPoint PPT Presentation

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  • Cathode raysHow are cathode rays produced in a discharge tube?What is the principle of thermionic emission?What is the work done on an electron accelerated through a p.d.?

  • Who invented the CRT?The cathode ray tube (CRT), invented by German physicist Karl Ferdinand Braun in 1897, is an evacuated glass envelope containing an electron gun (a source of electrons) and a fluorescent screen, with internal or external means to accelerate and deflect the electrons. When electrons strike the fluorescent screen, light is emitted.

    The electron beam is deflected and modulated in a way which causes it to display an image on the screen. The image may represent electrical waveforms (oscilloscope), pictures (old style television, computer monitor), echoes of aircraft detected by radar, etc.

  • QWCDescribe the principle of thermionic emission in an oscilloscope: (6 marks)Thermionic emission is a way of producing an electron beamMetals contain free (or conduction electrons)A metal is heated, some of the electrons gain sufficient kinetic energy to leave the metal at its surface. In practice, the metal is a wire filament which is heated by passing an electric current through it. The filament or .cathode. is at one end of an evacuated glass tube with a metal plate or .anode. At the other

  • Work done moving through a p.d. The electrons leave the cathode with negligible speed. They are accelerated by the attractive force of the positively charged anode. By the time they leave the gun, the electrons have energy eV, where e is the charge on the electron (1.6 10-19 C) and V is the anode voltage.

    Energy = charge voltage = eVAll the energy in the electron is kinetic, so we can say:Kinetic energy = mv2So we can combine the equations to give:eV = mv2

    Mass of an electron is 9.11 10-31 kgCalculate the kinetic energy and speed of an electron where the anode voltage is (a) 400 V(b) 400 kV

  • (a) Ek = eV = 1.6 10-19 400 V = 6.4 10-17J v2 = eEk/m = 2 6.4 10-17J 9.11 10-31 kg = 1.41 1014 m2 s-2 v = 1.18 107 m/s (b) Ek = eV = 1.6 10-19 400 000 V = 6.4 10-14J v2 = eEk/m = 2 6.4 10-14J 9.11 10-31 kg = 1.41 1017 m2 s-2 v = 3.74 108 m/s

  • Use the data sheet to calculate the following:The speed of an alpha particle accelerated across 300VThe speed of a lithium ion accelerated across 50kVThe speed of a chloride ion accelerated across 10kVA calcium ion accelerated through 3kV

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